91Ó°ÊÓ

Kinetic Energy is the energy possessed by an object due to its motion. It depends on the mass and velocity of the object and is given by the formula: \[ KE = \frac{1}{2} mv^2 \]where \( KE \) is the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. This formula shows that the kinetic energy increases with the square of the velocity. Thus, doubling the velocity will quadruple the kinetic energy, provided the mass remains constant.In the context of our problem, the particle's kinetic energy changes as it moves from one position to another under the influence of the force acting on it. Initially, it has \( 20.0 \, \mathrm{J} \) of kinetic energy, and this decreases to \( 11.0 \, \mathrm{J} \) as the particle moves to \( x = 3.00 \, \mathrm{m} \). This decrease in kinetic energy indicates that energy is being transferred from the particle to the surroundings, mainly due to the work done by the force acting against the particle's motion.

Work Done by a Force is a measure of energy transfer when a force moves an object through a distance along the direction of the force. It is calculated using the formula: \[ W = F \cdot d \cdot \cos \theta \]where \( F \) is the magnitude of the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement. When the force is applied in the direction of motion, the work done is positive, leading to an increase in the object's kinetic energy.In scenarios where multiple forces are involved, or the force varies with position, it is often more complex. For our problem, the work done is due to a non-constant force \( \vec{F} = (c x - 3.00 x^2) \hat{i} \). This required using an integral to determine the work over the path from \( x = 0 \) to \( x = 3.00 \, \mathrm{m} \). Ultimately, this work equates to a change in the particle's kinetic energy, according to the work-energy principle.

The Integral of Force is used to calculate the work done by a force that varies along a path, rather than being constant. This involves integrating the force function across the object's path of motion.For our specific problem, the force is given as \( F(x) = cx - 3.00x^2 \). Therefore, the work done as the particle moves from point \( x = 0 \) to \( x = 3.00 \, \mathrm{m} \) is expressed by:\[ W = \int_{0}^{3.00} (cx - 3.00x^2) \, dx \]By solving the integral:\[ \int cx \, dx = \frac{cx^2}{2} \quad \text{and} \quad \int 3.00x^2 \, dx = \frac{3.00x^3}{3} \]we arrive at the expression\[ \frac{9c}{2} - 27 \]This demonstrates how integral calculus can convert a position-dependent force into a specific amount of work done over that range, aiding in finding the unknown constant \( c \) in our physics problem.

Change in Kinetic Energy refers to how the energy of a particle or an object varies as its speed changes due to work done on or by it. It can be expressed by the work-energy principle, which states:\[ \Delta KE = KE_f - KE_i \]where \( \Delta KE \) is the change in kinetic energy, \( KE_f \) is the final kinetic energy, and \( KE_i \) is the initial kinetic energy. In the problem, the kinetic energy transitions from an initial value of \( 20.0 \, \mathrm{J} \) to a final value of \( 11.0 \, \mathrm{J} \), resulting in a change of \(-9.0 \, \mathrm{J} \). This negative value indicates that the particle has lost energy, which aligns with the work calculated being done on the particle by an external force. This knowledge, combined with the work done derived from integrating the force, enabled us to solve for \( c \) — the constant in the force equation.